bio | website | |
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location | CMLS, X | |
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visits | member for | 3 years, 1 month |
seen | Jun 13 at 22:38 | |
stats | profile views | 1,332 |
I'm working in additive theory, mainly in the context of non-abelian [semi]groups. I got a PhD in pure mathematics in Nov 2013 from the Université Lyon 1, under the advisorship of François Hennecart and Alain Plagne. As from Jan 2014, I am a post-doc at the Centre de mathématiques Laurent Schwartz.
Jul 18 |
awarded | Yearling |
Jul 2 |
awarded | Inquisitive |
Jul 2 |
awarded | Curious |
Apr 30 |
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An introduction to sieve method and their application, Cojocaru & Murty
You should definitely replace the '$x$' appearing as an upper bound for the variable $\delta$ in the expression of the integral $I$ with '$t$'. |
Apr 23 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
I'm finally convinced: there're first-countable topologies which are not semimetrizable (see the comments to mathoverflow.net/questions/163559 for details). |
Apr 23 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
[...] About the origins of basic ideas in the area of asymmetric topology'' (in C. E. Aull and R. Lowen (eds.), Handbook of the History of General Topology, Vol. 3, Dordrecht: Kluwer (2001), 853-968), reports a letter by the same Fox where even a paracompact Hausdorff counterexample (to the $\gamma$-space conjecture) is provided. |
Apr 23 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
[...] (see Distance function and the metrization problem, BAMS 43 (1937), 133-142). However, it is still true that not all first-countable topologies are semimetrizable, and I learned from R. Fox' work that the question is related to the $\gamma$-space problem (every semimetrizable space is a $\gamma$-space, and it took some time before a disproof of the converse). A Hausdorff counterexample is, in fact, given in: R. Fox, Solution of the $\gamma$-space problem, Proc. AMS 85 (1982), 606-608. And H.-P. A. Künzi, in his survey "Nonsymmetric distances and their associated topologies: [...] |
Apr 23 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
Errata corrige. My "proof" that $(X,\tau)$ isn't semimetrizable when $X$ is countably infinite and $\tau$ is the cofinite topology on $X$ was flawed, and in fact, the contrary is true! This follows from a (straightforward) generalization of a theorem by W. A. Wilson dating back to the 1930s, which appears, e.g., as Theorem 6.3.50 in J. Goubault-Larrecq, Non-Hausdorff Topology and Domain Theory: Selected Topics in Point-Set Topology (New Math. Monographs 22, Cambridge Univ. Press, 2013), or can be recovered as an instance of a theorem by A. H. Frink on countably-based quasi-uniformities [...] |
Apr 20 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
I'm considering monoids (to me, a monoid is a unital semigroup, or, if you prefer, a unital associative magma in the sense of Bourbaki). But I'm not sure to get the point of your question! What do you mean? Btw, my use of "in addition" in the present formulation of Q2 is misleading: every cancellative monoid is resilient. |
Apr 20 |
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Embedding a linearly ordered free monoid into a linearly ordered group
I will try to dig into all of this, but thanks for sharing your insights! A minor note: I wouldn't say that "bi-ordered" is more conventional than "linearly ordered". The latter appears, e.g., in K. Iwasawa's and P. Conrad's papers. The same Conrad coined (?) the term "o-group" in Extensions of Ordered Groups (Proc. AMS, Vol. 6, No. 4 (Aug., 1955), pp. 516-528), which is still in use by a number of authors. Others do simply talk of "ordered groups" (e.g., G. Freiman and coauthors in the additive theory of groups), and the list continues. But yes, some people prefer "bi-ordered". |
Apr 20 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
Sorry for the late reply (I was on holiday). In fact, there're 1st-countable topologies which are not semimetrizable (to be honest, I had no doubt about their existence, but it's only two days ago that I found a counterexample): this is the case, e.g., with the cofinite topology on a countably infinite set. So yes, I restated the OP to take into account your comments. |
Apr 20 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
Restated the question to take into account Eric and Chris' comments |
Apr 16 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
@Chris: Thanks for your clarifications, which change many things and prove that also my 2nd reading of your answer was incorrect. To be honest, it wasn't so clear (to me?) which open sets you were actually adding to $\tau_X$: there's not a unique way to do it, and the "cheapest" way to do it is to add just $M$. Further, your answer still says: "Let $X$ be any topological space. [...]" So it's not really a question of being complicated or not. Btw, there's a mistake in my last comment: it should really read as "$d(0,x):=d(1,x):=0$ for all $x\in M$ and $d(x,0):=d(x,1):=\infty$ for $x\in X$". |
Apr 16 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
Here's a "natural" follow up: mathoverflow.net/questions/163559. |
Apr 16 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
added 67 characters in body |
Apr 16 |
asked | First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric |
Apr 16 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
If my 2nd reading of this answer (based on Eric's comments) is right, then Chris' construction doesn't however work, at least in general (and it's not clear to me if it works at all). Eric is in fact right: a semimetric $d_X$ on $X$ always extends to a semimetric $d$ on $M$ by $d(0,x):=d(x,0):=d(1,y):=d(y,1):=\infty$ for all $x,y\in M$ with $x\ne 0$ and $y\ne 1$; but the only neighborhood of both $0$ and $1$ in the canonical topology of $d$ is $M$, so $d$ induces $\tau_X\cup\{M\}$ if $d_X$ induces $\tau_X$ (the topology initially given on $X$). The rest follows from the previous comment. |
Apr 16 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
@Eric Wofsey. I see! So I was misreading Chris' answer. He's just starting with a topology $\tau_X$ on $X$ and taking $\tau=\tau_X\cup\{M\}$ as a topology on $M$, right? That's fine! Still, how to prove that $\tau$ is not induced by a left/right $\mathbb M$-invariant semimetric? This boils down to $\tau$ not being semimetrizable at all (which is close to your remarks), since for a semimetric $d$ on $M$ we would have (for Chris' construction) that, for all $x,y,z\in M$, $d(xz,yz)=d(zx,zy)=d(0,0)\le d(x,y)$ if $z\ne 1$ or $d(xz,yz)=d(zx,zy)=d(x,y)$ if $z=1$, i.e. $d$ would be subinvariant. |
Apr 16 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
(...) for which $\mathbb M$ is the unitization of a null sgrp, how do you think to prove that there doesn't exist any left/right $\mathbb M$-subinvariant semimetric $d$ on $M$ such that $\tau$ the canonical topology induced by $d$? That's the thing. |
Apr 16 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
(...) $(\mathbb M,\tau)$ is not a topological monoid, unless $\{S\}\in\tau$, which is the case iff $S=\{0\}$. In fact, $M\setminus \{0\}$ is an open neighborhood of $1$ in $\tau$ (by construction), and given any open neighborhood $U$ of $(S,S)$ in the product topology induced on $M\times M$ by $\tau$ we should have that $xy\ne 0$ for all $x,y\in U$, but this is false unless we can take $U=\{(S,S)\}$, which in turn is possible iff $M\setminus \{0\}=\{S\}$, i.e. $S=\{0\}$. Do I miss something? And in any case, even assuming that $(\mathbb M,\tau)$ is a 1st-countable topological monoid (...) |